Advanced Calculus Single Variable

6.8 Exercises

A function f is Lipschitz continuous or just Lipschitz for short if there exists a constant,
K such that

|f (x)− f (y)| ≤ K |x − y|

for all x,y ∈ D. Show every Lipschitz function is uniformly continuous.

If

|xn − yn|

→ 0 and xn→ z, show that yn→ z also. This was used in the proof of
Theorem 6.7.2.

Consider f :

(1,∞)

→ ℝ given by f

(x)

=

1x

. Show f is uniformly continuous even
though the set on which f is defined is not sequentially compact.

If f is uniformly continuous, does it follow that

|f|

is also uniformly continuous? If

|f|

is uniformly continuous does it follow that f is uniformly continuous? Answer the same
questions with “uniformly continuous” replaced with “continuous”. Explain
why.

Suppose f is a function defined on D and λ ≡ inf

{f (x) : x ∈ D}

. A sequence

{x }
n

of
points of D is called a minimizing sequence if limn→∞f

(x )
n

= λ. A maximizing
sequence is defined analogously. Show that minimizing sequences and maximizing
sequences always exist. Now let K be a sequentially compact set and f : K → ℝ. Show
that f achieves both its maximum and its minimum on K by considering directly
minimizing and maximizing sequences. Hint: Let M ≡ sup

{f (x) : x ∈ K }

. Argue
there exists a sequence,

{x }
n

⊆ K such that f

(x )
n

→ M. Now use sequential
compactness to get a subsequence,

{x }
nk

such that limk→∞xnk = x ∈ K and use the
continuity of f to verify that f

(x)

= M. Incidentally, this shows f is bounded on
K as well. A similar argument works to give the part about achieving the
minimum.

Let f : D → ℝ be a function. This function is said to be lower
semicontinuous3
at x ∈ D if for any sequence

{x }
n

⊆ D which converges to x it follows

f (x) ≤ lim inf f (xn).
n→ ∞

Suppose D is sequentially compact and f is lower semicontinuous at every point of D.
Show that then f achieves its minimum on D.

Let f : D → ℝ be a function. This function is said to be upper semicontinuous at x ∈ D
if for any sequence

{xn }

⊆ D which converges to x it follows

f (x) ≥ lim ns→up∞ f (xn).

Suppose D is sequentially compact and f is upper semicontinuous at every point of D.
Show that then f achieves its maximum on D.

Show that a real valued function is continuous if and only if it is both upper and lower
semicontinuous.

Give an example of a lower semicontinuous function which is not continuous and an
example of an upper semicontinuous function which is not continuous.

Suppose

{f : α ∈ Λ}
α

is a collection of continuous functions. Let

F (x) ≡ inf{fα(x) : α ∈ Λ}

Show F is an upper semicontinuous function. Next let

G (x) ≡ sup{fα(x) : α ∈ Λ}

Show G is a lower semicontinuous function.

Let f be a function. epi

(f)

is defined as

{(x,y) : y ≥ f (x)}.

It is called the epigraph of f. We say epi

(f)

is closed if whenever

(xn,yn)

∈epi

(f)

and
xn→ x and yn→ y, it follows

(x,y)

∈epi

(f)

. Show f is lower semicontinuous if and
only if epi

(f)

is closed. What would be the corresponding result equivalent to upper
semicontinuous?